A proof of a conjecture by Jabara on groups in which all involutions are odd transpositions

Let 𝐺 be a group, let Γ 2 be the set of elements of order 2 in 𝐺, and assume that

1. G = ⟨ Γ 2 ⟩ ,

2. if a , b ∈ Γ 2 , then the order of a ⁢ b is 1, 2, 3 or 5 and all these values actually occur.

Let 𝐺 be a group which satisfies hypotheses (H.1) and (H.2).

1. If Γ 4 = ∅ , then G ≃ A 5 .

2. If Γ 4 ≠ ∅ , then G ≃ PSU ⁢ ( 3 , 4 ) .

Let 𝐺 be a group which satisfies hypotheses (H.1) and (H.2), and let 𝐻 be a subgroup of 𝐺. Then ϖ ⁢ ( Π ⁢ ( H ) ) ⊆ { 1 , 2 , 3 , 5 } , but clearly, it can happen that H ≠ ⟨ Γ 2 ⁢ ( H ) ⟩ (and also that Γ 2 ⁢ ( H ) = ∅ ).

In principle, nothing can be said about quotients of 𝐺.

Let G ≃ PSU ⁢ ( 3 , 4 ) , and let 𝑆 be a Sylow 2-subgroup of 𝐺.

(A) Then 𝐺 has a unique conjugacy class of involutions, i.e. Γ 2 = a G ( a ∈ Γ 2 ).

(B) 𝑆 has order 2 6 and class 2, ⟨ Γ 2 ⁢ ( S ) ⟩ = Z ⁢ ( S ) ≃ C 2 × C 2 , and 𝑆 is a Suzuki 2-group in the sense of Higman [10]. Moreover, N G ⁢ ( S ) ≃ S : C 15 , and its Hall { 2 , 3 } -subgroup L 192 = S : C 3 is a Frobenius group of order 192. The Frobenius group L 48 = ( C 4 × C 4 ) : C 3 (of order 48) is a subgroup of L 192 . (We want to point out that there are two non-isomorphic Frobenius groups of order 48 and five non-isomorphic Frobenius groups of order 192.)

(C) The maximal subgroups of 𝐺 are

(D) Let H , K ≤ G be such that H ≃ A 5 ≃ K and H ≠ K . Then H ∩ K = 1 or H ∩ K ≃ C 2 × C 2 . Moreover, if H ∩ K ≃ C 2 × C 2 , then

has the presentations ⟨ a , x , y ∣ a 2 , x 3 , y 3 , ( a ⁢ x ) 3 , ( a ⁢ y ) 3 , a ⁢ ( x ⁢ y - 1 ) 2 ⟩ , or, more simply,

(E) Let L 192 = S : C 3 , let L 48 ≃ ⟨ x , y ⟩ as in ( ★ D ), and let a = ( x ⁢ y - 1 ) 2 . If L 48 < L 192 , then there is some z ∈ Γ 3 ⁢ ( L 192 ) such that L 192 has the presentation

Let a , b , c ∈ Γ 2 be such that [ a , b ] = 1 , [ b , c ] ≠ 1 , a ≠ b , a ≠ c . Then ⟨ a , b , c ⟩ ≃ A 5 and ( o ⁢ ( b ⁢ c ) , o ⁢ ( c ⁢ a ) , o ⁢ ( a ⁢ b ⁢ c ) ) ∈ { ( 3 , 5 , 5 ) , ( 5 , 3 , 5 ) , ( 5 , 5 , 3 ) } .

Proof

Taking into account that a ⁢ b ∈ Γ 2 , we can consider the groups

with ℓ ∈ { 3 , 5 } and i 1 , i 2 ∈ { 2 , 3 , 5 } . A computation with GAP shows that

Moreover, J ⁢ ( 5 , 5 , 5 ) ≃ PSL ⁢ ( 2 , 11 ) , and this case must be excluded because D 12 is a subgroup of PSL ⁢ ( 2 , 11 ) . For the other values of ℓ, i 1 and i 2 , the group 𝐽 is isomorphic to a group in the set { 1 , C 2 , C 2 × C 2 , D 6 , D 10 } . If J ∈ { 1 , C 2 , D 6 , D 10 } , then 𝐽 does not contain two distinct involutions which commute. If J ≃ C 2 × C 2 , then 𝐽 does not contain two distinct involutions which do not commute. In both cases, the hypotheses of the lemma are not satisfied, and therefore we obtain a contradiction. So the statement is proved. ∎

𝐺 contains a unique conjugate class of involutions.

Proof

We begin by observing that Γ 2 ⁢ ( Z ⁢ ( G ) ) = ∅ : assume, arguing by contradiction, a ∈ Z ⁢ ( G ) ∩ Γ 2 , and let b , c ∈ Γ 2 be such that o ⁢ ( b ⁢ c ) = 3 (or o ⁢ ( b ⁢ c ) = 5 ). Since a ∈ Z ⁢ ( G ) , we have a ⁢ b ∈ Z ⁢ ( G ) , and ( a ⁢ b ) ⁢ c has order 6 (or 10), a contradiction.

Let a , b ∈ Γ 2 , a ≠ b . If [ a , b ] ≠ 1 , then ⟨ a , b ⟩ is isomorphic to D 6 or to D 10 , so 𝑏 is conjugate to 𝑎. If [ a , b ] = 1 , then (by (H.1)) we can choose c ∈ Γ 2 such that [ b , c ] ≠ 1 . From Lemma 3.1, ⟨ a , b , c ⟩ ≃ A 5 , and 𝑏 is conjugate to 𝑎 in ⟨ a , b , c ⟩ . ∎

𝐺 has no subgroups isomorphic to C 2 × C 2 × C 2 . In particular, if a ∈ Γ 2 , then E a ≃ C 2 × C 2 and C G ⁢ ( a ) ≤ N G ⁢ ( E a ) .

Proof

Let a , b , c ∈ Γ 2 be such that ⟨ a , b , c ⟩ ≃ C 2 × C 2 × C 2 , and let t ∈ Γ 2 be such that a ⁢ t ∈ Γ 3 . By Lemma 3.1, we obtain

and b ⁢ t , c ⁢ t , b ⁢ c ⁢ t ∈ Γ 5 . Since [ b , t ] ≠ 1 ≠ [ c , t ] , we have also ⟨ b , c , t ⟩ ≃ A 5 , and hence b ⁢ t or c ⁢ t or b ⁢ c ⁢ t has order 3. Assume b ⁢ t ∈ Γ 3 . Then ( b ⁢ t ) 5 = 1 = ( b ⁢ t ) 3 implies b ⁢ t = 1 and t = b , a contradiction.

Hence if b , c ∈ C G ⁢ ( a ) ∩ Γ 2 , then [ b , c ] ≠ 1 . But

which is impossible. So C G ⁢ ( a ) ∩ Γ 2 = { a , b , a ⁢ b } and E a = ⟨ a , b ⟩ ≃ C 2 × C 2 .

Let g ∈ C G ⁢ ( a ) , and suppose E a = ⟨ a , b ⟩ . Then we have [ a , b g ] = 1 , b g ∈ E a and g ∈ N G ⁢ ( E a ) . ∎

Let H ≤ G be such that H ≃ A 5 or H ≃ PSU ⁢ ( 3 , 4 ) . If a ∈ Γ 2 ⁢ ( H ) , then E a ≤ H .

Proof

The claim is an easy consequence of Lemma 3.3. ∎

Let a , b ∈ Γ 2 be such that [ a , b ] ≠ 1 . Then ⟨ E a , E b ⟩ ≃ A 5 .

Proof

Let c ∈ E b with 1 ≠ c ≠ b . Then, by Lemma 3.1, H = ⟨ a , b , c ⟩ ≃ A 5 . By Lemma 3.4, we have E a ≤ H and E b ≤ H , and hence we can conclude that ⟨ E a , E b ⟩ ≃ A 5 . ∎

Let 𝑇 be a group such that ϖ ⁢ ( Π ⁢ ( T ) ) ⊆ { 1 , 2 , 3 , 5 } , let 𝑁 be a normal subgroup of 𝑇, and assume that 𝑁 is periodic and Γ 2 ⁢ ( N ) = ∅ . Then we have ϖ ⁢ ( Π ⁢ ( T / N ) ) = ϖ ⁢ ( Π ⁢ ( T ) ) ; in particular, ϖ ⁢ ( Π ⁢ ( T / N ) ) ⊆ { 1 , 2 , 3 , 5 } .

Proof

Let a ⁢ N , b ⁢ N ∈ Γ 2 ⁢ ( T / N ) . Since 𝑁 is periodic and without involution, we can assume a , b ∈ Γ 2 ⁢ ( T ) and ( a ⁢ b ) k = 1 for some k ∈ { 2 , 3 , 5 } . Then we have ( a ⁢ N ⁢ b ⁢ N ) k = ( a ⁢ b ⁢ N ) k = N . ∎

𝐺 does not contain subgroups isomorphic to L 18 or to W 54 .

Proof

We argue by contradiction. Assume that L 18 < G , and let a , b , c ∈ Γ 2 be such that L 18 ≃ ⟨ a , b , c ∣ ρ L ⟩ is as in (3.1). Let t ∈ Γ 2 ∖ { a } be such that [ a , t ] = 1 , and let

For every δ ∈ Δ 0 , we have a ⁢ δ ∈ Γ 3 , and hence, by Lemma 3.1, t ⁢ δ , a ⁢ t ⁢ δ ∈ Γ 5 . Let

and define

A computation with GAP shows that J ⁢ ( 2 ) = J ⁢ ( 3 ) = J ⁢ ( 5 ) = 1 , a contradiction.

Assume that W 54 < G , and let a , b , c ∈ Γ 2 be such that W 54 = ⟨ a , b , c ⟩ is as in (3.2). Let t ∈ Γ 2 ∖ { a } be such that [ a , t ] = 1 , and let z = ( a ⁢ b ⁢ c ) 2 ∈ Z ⁢ ( W ) . By Lemma 3.3, t z ∈ E a , so t z = t or t z = a ⁢ t , and since 𝑧 has order 3, we must have t z = t , so z ∈ Z ⁢ ( ⟨ a , b , c , t ⟩ ) .

Let S ¯ = ⟨ a ¯ , b ¯ , c ¯ , t ¯ ⟩ = ⟨ a , b , c , t ⟩ / ⟨ z ⟩ , R ¯ = ⟨ a ¯ , b ¯ , c ¯ ⟩ ≃ L 18 , and let 𝑆 and 𝑅 be the complete preimage of R ¯ and S ¯ in 𝐺. By Lemma 3.6, ϖ ⁢ ( Π ⁢ ( S ¯ ) ) ⊆ { 1 , 2 , 3 , 5 } , and since a ⁢ t ⁢ b ∈ Γ 5 ⁢ ( Π ⁢ ( S ¯ ) ) , ϖ ⁢ ( Π ⁢ ( S ¯ ) ) = { 1 , 2 , 3 , 5 } . So S ¯ satisfies hypotheses (H.1) and (H.2) and R ¯ < S ¯ with R ¯ ≃ L 18 . This is in contradiction with what we have shown in the first part of the proof. ∎

Let x ∈ Π ∩ Γ 3 . Assume that x = a ⁢ b with a , b ∈ Γ 2 . Then we have { t ∈ Γ 2 ∣ x t = x - 1 } = { a , b , a b } .

Proof

Let t ∈ Γ 2 be such that x t = x - 1 , assume t ∉ { a , b , a b } , and let

An easy computation shows that J ⁢ ( 2 ) ≃ D 12 and J ⁢ ( 5 ) ≃ D 30 , and we must rule out such cases by (H.2), while J ⁢ ( 3 ) ≃ ( C 3 × C 3 ) : C 2 = L 18 , against Lemma 3.7. ∎

Let

If the subgroup H = ⟨ a , b , c ⟩ cannot be generated by less than three involutions, then 𝐻 is said to be a subgroup of type ( o ⁢ ( a ⁢ b ) , o ⁢ ( b ⁢ c ) , o ⁢ ( c ⁢ a ) ) .

If H ≤ G is of type ( 3 , 3 , 3 ) , then H ≃ L 150 .

Proof

Let

with i 1 , i 2 ∈ { 2 , 3 , 5 } . A computation shows that J ⁢ ( 2 , 3 ) ≃ S 4 , which is impossible, J ⁢ ( 3 , 3 ) ≃ E 27 : C 2 , which is excluded for Lemma 3.7 and J ⁢ ( 5 , 3 ) ≃ L 150 . Since for other values of i 1 and i 2 , we have | J | ≤ 6 , the lemma is proved. ∎

If H < G is of type ( 3 , 3 , 5 ) , then 𝐻 can be isomorphic only to A 5 , to L 150 or to PSU ⁢ ( 3 , 4 ) .

Proof

Let

with i 1 , i 2 ∈ { 2 , 3 , 5 } .

A computation shows that J ⁢ ( 5 , 2 ) ≃ C 2 × A 5 with Z ⁢ ( J ) = ⟨ ( a ⁢ b ⁢ c ) 3 ⟩ , so in this case, we have ⟨ a , b , c ⟩ ≃ A 5 , J ⁢ ( 3 , 3 ) ≃ L 150 and J ⁢ ( 5 , 5 ) ≃ C 2 × PSU ⁢ ( 3 , 4 ) with Z ⁢ ( J ) = ⟨ ( a ⁢ b ⁢ c ) 15 ⟩ , so in this case, ⟨ a , b , c ⟩ ≃ PSU ⁢ ( 3 , 4 ) . Since for other values of i 1 and i 2 , we have | J | ≤ 2 , the lemma is proved. ∎

If H < G is such that H ≃ L 150 , then for every a ∈ Γ 2 ⁢ ( H ) , we have ⟨ H , E a ⟩ ≃ PSU ⁢ ( 3 , 4 ) .

Proof

Let H = ⟨ a , b , c ∣ ρ ⟩ ≃ L 150 , where

Then w = ( c ⁢ a ⁢ b ) 2 ∈ Γ 5 ⁢ ( H ) and [ a , w ] = 1 . Let t ∈ Γ 2 be such that E a = ⟨ a , t ⟩ . By Lemma 3.3, t w ∈ E a , and since 𝑤 has odd order, [ t , w ] = 1 .

If h = b or h = c , then, by Lemma 3.1, we have t ⁢ h , a ⁢ t ⁢ h ∈ Γ 5 . Since a ⁢ b c ∈ Γ 5 again by Lemma 3.1, we have

1. t ⁢ b c ∈ Γ 3 and a ⁢ t ⁢ b c ∈ Γ 5 or

2. t ⁢ b c ∈ Γ 5 and a ⁢ t ⁢ b c ∈ Γ 3 .

Let σ = { t 2 , ( a ⁢ t ) 2 , [ t , w ] , ( t ⁢ b ) 5 , ( a ⁢ t ⁢ b ) 5 , ( t ⁢ c ) 5 , ( a ⁢ t ⁢ c ) 5 } , and consider the groups

A computation with GAP shows that J ⁢ ( 3 ) ≃ PSU ⁢ ( 3 , 4 ) ≃ J ⁢ ( 5 ) . ∎

Three involutions a , b , c ∈ Γ 2 are said to be independent if

Let a , b , c ∈ Γ 2 be three independent involutions of 𝐺. Then we have ⟨ E a , E b , E c ⟩ ≃ PSU ⁢ ( 3 , 4 ) .

Proof

Let E a = ⟨ a , α ⟩ , E b = ⟨ b , β ⟩ and E c = ⟨ c , γ ⟩ , and let J = ⟨ E a , E b , E c ⟩ . Fix 𝑎. Then, by Lemma 3.1, there are b 1 ∈ E b and c 1 ∈ E c such that a ⁢ b 1 , a ⁢ c 1 ∈ Γ 3 . We can, without loss of generality, assume that b = b 1 and c = c 1 , and hence H = ⟨ a , b , c ⟩ is of type ( 3 , 3 , 3 ) or ( 3 , 3 , 5 ) . From Lemmas 4.2 and 4.3, we must consider the following three cases.

(1) H ≃ L 150 . If J 1 = ⟨ H , E a ⟩ , J 2 = ⟨ H , E b ⟩ and J 3 = ⟨ H , E a ⟩ , then, by Lemma 4.4, we can conclude that J 1 ≃ J 2 ≃ J 3 ≃ PSU ⁢ ( 3 , 4 ) and, by Lemma 3.4, E a ∪ E b ∪ E c ⊆ J 1 ∩ J 2 ∩ J 3 .

Hence J = J 1 = J 2 = J 3 ≃ PSU ⁢ ( 3 , 4 ) .

(2) H ≃ A 5 , by Lemma 3.5, H = ⟨ E a , E b ⟩ , and by Lemma 3.4, E c ≤ H , against the hypotheses. So this case is not possible.

(3) H ≃ PSU ⁢ ( 3 , 4 ) , and by Lemma 3.4, E a , E b , E c ≤ H .

Hence J = H ≃ PSU ⁢ ( 3 , 4 ) . ∎

Let x ∈ Γ 3 . Then there is an element d ∈ Γ 2 such that x d = x - 1 or x ∈ Z ⁢ ( G ) .

Proof

If [ a , x ] = 1 for every a ∈ Γ 2 , then we should have [ x , ⟨ Γ 2 ⟩ ] = 1 and x ∈ Z ⁢ ( G ) , since G = ⟨ Γ 2 ⟩ . If [ a , a x ] = 1 for every a ∈ Γ 2 , then ⟨ a , x ⟩ ≃ A 4 and ( a ⁢ x ) 3 = 1 for every a ∈ Γ 2 . Let b ∈ Γ 2 be such that a ⁢ b ∈ Γ 3 . Taking into account that a x ⁢ b ∈ Γ 5 , with GAP, we can verify that the group

has order 3 (and, in 𝐻, a = b ), which is impossible.

Let a ∈ Γ 2 be such that [ a , x ] ≠ 1 , let b = a x , c = b x and J = ⟨ E a , E b , E c ⟩ . By Lemma 4.6, 𝐽 is finite, and since we can easily verify that E a x = E b and E b x = E c , we have J = J x . Since E a x ≠ E c , we must consider the following two cases.

(1) J ≃ A 5 . Since Aut ⁢ ( J ) ≃ A 5 : C 2 , we have x ∈ J , and we conclude because every element of Γ 3 ⁢ ( A 5 ) is inverted by some involution in A 5 .

(2) J ≃ PSU ⁢ ( 3 , 4 ) . Since Aut ⁢ ( J ) ≃ PSU ⁢ ( 3 , 4 ) : C 4 , we have x ∈ J , and we conclude because every element of Γ 3 ⁢ ( PSU ⁢ ( 3 , 4 ) ) is inverted by some involution in PSU ⁢ ( 3 , 4 ) . ∎

If Γ 4 = ∅ , then G ≃ A 5 .

Proof

Let H ≤ G with H ≃ A 5 , and let a , b ∈ Γ 2 ⁢ ( H ) be such that E a ≠ E b . From Lemma 3.5, we have H = ⟨ E a , E b ⟩ . Let c ∈ Γ 2 be such that c ∉ H . Then, by Lemma 3.4, H ∩ E c = 1 , and by Lemma 4.6, ⟨ E a , E b , E c ⟩ ≃ PSU ⁢ ( 3 , 4 ) .

Since PSU ⁢ ( 3 , 4 ) contains elements of order 4, this forces Γ 2 ⊆ H , and by (H.1), G = ⟨ Γ 2 ⟩ , so G = H ≃ A 5 . ∎

If Γ 4 ≠ ∅ , then there is H ≤ G such that H ≃ PSU ⁢ ( 3 , 4 ) .

Proof

Let a , b ∈ Γ 2 with a ⁢ b ≠ b ⁢ a . Then ⟨ E a , E b ⟩ ≃ A 5 . Since Γ 4 ⁢ ( A 5 ) = ∅ and G = ⟨ Γ 2 ⟩ , there is c ∈ Γ 2 such that c ∉ ⟨ E a , E b ⟩ . From Lemma 4.6, we have H = ⟨ E a , E b , E c ⟩ ≃ PSU ⁢ ( 3 , 4 ) . ∎

Let E a ≤ G and x , y ∈ Γ 3 be such that ⟨ E a , x ⟩ ≃ A 4 ≃ ⟨ E a , y ⟩ . If ⟨ E a , x ⟩ ≠ ⟨ E a , y ⟩ , then

1. ⟨ E a , x , y ⟩ = ⟨ x , y ⟩ ≃ ( C 4 × C 4 ) : C 3 = L 48 , and we have

   (5.1) ⟨ x , y ⟩ = ⟨ a , x , y ∣ x 3 , y 3 , ( x ⁢ y ) 3 , ( x ⁢ y - 1 ) 4 ⟩

   or

   (5.2) ⟨ x , y ⟩ = ⟨ a , x , y ∣ x 3 , y 3 , ( x ⁢ y ) 4 , ( x ⁢ y - 1 ) 3 ⟩ .

   In particular, we can choose y * ∈ Γ 3 ⁢ ( L 48 ) so that a = ( x ⁢ y * - 1 ) 2 .

2. There is only one H ≤ G such that H ≃ PSU ⁢ ( 3 , 4 ) and ⟨ E a , x , y ⟩ < H .

Proof

Since x , y ∉ Z ⁢ ( G ) , by Lemma 4.7, we have that there are b , c ∈ Γ 2 such that x b = x - 1 and y c = y - 1 .

Let X = ⟨ E a , E b ⟩ ≃ A 5 and β = b ⁢ x ∈ Γ 2 . Let a 1 ∈ E a be such that a 1 ⁢ b ∈ Γ 3 , and let a 2 = a 1 x . Then E a = E a 1 and X = ⟨ E a 1 , E b ⟩ = ⟨ a 1 , a 2 , b ∣ ξ ⟩ , where

Let

A computation shows that T ≃ A 5 , and hence T = X , ⟨ E a , E b , E β ⟩ = ⟨ E a , E b ⟩ , so E β ≤ X and x ∈ X . In the same way, we get that y ∈ Y = ⟨ E a , E c ⟩ .